There are two machines available for cutting corks intended for use in bottles. The first produces corks with diameters that are normally distributed with mean 3 cm and standard deviation 0.1 cm. The second machine produces corks with diameters that have a normal distribution with mean 3.04 cm and standard deviation 0.02 cm. Acceptable corks have diameters between 2.9 cm and 3.1 cm. Which machine is more likely to produce an acceptable cork? What should the acceptable range for cork diameters be (from 3 − d cm to 3 + d cm) to be 90% certain for the first machine to produce an acceptable cork?

a) The second machine is more likely to produce an acceptable cork.

b) Acceptable range for cork diameters produced by the first machine with a 90% confidence = (2.8355, 3.1645)

Step-by-step explanation:

This is a normal distribution problem

For the first machine,

Mean = μ = 3 cm

Standard deviation = σ = 0.1 cm

And we want to find which percentage of the population falls between 2.9 cm and 3.1 cm.

P(2.9 ≤ x ≤ 3.1) = P(x ≤ 3.1) - P(x ≤ 2.9)

We first standardize this measurements.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 2.9 cm

z = (x - μ)/σ = (2.9 - 3.0)/0.1 = - 1.00

For 3.1 cm

z = (x - μ)/σ = (3.1 - 3.0)/0.1 = 1.00

P(x ≤ 3.1) = P(z ≤ 1.00) = 0.841

P(x ≤ 2.9) = P(z ≤ -1.00) = 0.159

P(2.9 ≤ x ≤ 3.1) = P(-1.00 ≤ z ≤ 1.00) = P(z ≤ 1.00) - P(z ≤ -1.00) = 0.841 - 0.159 = 0.682 = 68.2%

This means that 68.2% of the diameter of corks produced by the first machine lies between 2.9 cm and 3.1 cm.

For the second machine,

Mean = μ = 3.04 cm

Standard deviation = σ = 0.02 cm

And we want to find which percentage of the population falls between 2.9 cm and 3.1 cm.

P(2.9 ≤ x ≤ 3.1) = P(x ≤ 3.1) - P(x ≤ 2.9)

We standardize this measurements.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 2.9 cm

z = (x - μ)/σ = (2.9 - 3.04)/0.02 = - 7.00

For 3.1 cm

z = (x - μ)/σ = (3.1 - 3.0)/0.02 = 3.00

P(x ≤ 3.1) = P(z ≤ 3.00) = 0.999

P(x ≤ 2.9) = P(z ≤ -7.00) = 0.0

P(2.9 ≤ x ≤ 3.1) = P(-7.00 ≤ z ≤ 3.00) = P(z ≤ 3.00) - P(z ≤ -7.00) = 0.999 - 0.0 = 0.999 = 99.9%

This means that 99.9% of the diameter of corks produced by the second machine lies between 2.9 cm and 3.1 cm.

Hence, we can conclude that the second machine is more likely to produce an acceptable cork.

b) Margin of error = (z-multiplier) × (standard deviation of the population)

For 90% confidence interval, z-multiplier = 1.645 (from literature and the z-tables)

Standard deviation for first machine = 0.1

Margin of error, d = 1.645 × 0.1 = 0.1645.

The acceptable range = (mean ± margin of error)

Mean = 3

Margin of error = 0.1645

Lower limit of the acceptable range = 3 - d = 3 - 0.1645 = 2.8355

Upper limit of the acceptable range = 3 + d = 3 + 0.1645 = 3.1645

Acceptable range = (2.8355, 3.1645)